Let $h(x)=x^3e^{x}$. What is the absolute minimum value of $h$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{3e}$ (Choice B) B $-\dfrac{27}{e^3}$ (Choice C) C $-\dfrac{e^3}{9}$ (Choice D) D $h$ has no minimum value
Let's first find the relative extremum points of $h$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $h$. The derivative of $h$ is $h'(x)=x^2e^{x}(x+3)$. $h'(x)=0$ for $x=0,-3$. $h'$ is defined for all real numbers. Therefore, our only critical points are $x=0,-3$. Our critical points divide the function's domain (which is all real numbers) into three intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $x< \llap{-}3$ $\llap{-}3<x<0$ $x>0$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<-3$ $x=-4$ $h'(-4)=-\dfrac{16}{e^4}<0$ $h$ is decreasing $\searrow$ $-3<x<0$ $x=-1$ $h'\left(-1\right)=\dfrac2e>0$ $h$ is increasing $\nearrow$ $x>0$ $x=1$ $h'(1)=4e>0$ $h$ is increasing $\nearrow$ Let's imagine ourselves walking on the graph of $h$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down and down until we reach $x=-3$. Then, we will be forever going up. Therefore, $h$ must obtain its absolute minimum value at $x=-3$. We are asked to find that minimum value, which is $h\left(-3\right)=-\dfrac{27}{e^3}$. In conclusion, the absolute minimum value of $h$ is $-\dfrac{27}{e^3}$.